All possible two-factor products are formed f | vedclass

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Question

(B) The total number of ways to choose two distinct numbers from the set ${1, 2, 3, \dots, 200}$ is given by $^{200}C_2 = \frac{200 	imes 199}{2} = 1

Vedclass · Accepted Answer

$7180$

Vedclass · Answer

(B) The total number of ways to choose two distinct numbers from the set ${1, 2, 3, \dots, 200}$ is given by $^{200}C_2 = \frac{200 	imes 199}{2} = 19900$.$A$ product of two numbers is a multiple of $5$ if at least one of the numbers is a multiple of $5$.It is easier to calculate the complement: the number of products that are $NOT$ multiples of $5$.$A$ product is not a multiple of $5$ if both chosen numbers are not multiples of $5$.The numbers from $1$ to $200$ that are multiples of $5$ are $5, 10, \dots, 200$. The count is $\frac{200}{5} = 40$.The numbers that are $NOT$ multiples of $5$ are $200 - 40 = 160$.The number of ways to choose two numbers such that neither is a multiple of $5$ is $^{160}C_2 = \frac{160 	imes 159}{2} = 80 	imes 159 = 12720$.Therefore, the number of products that are multiples of $5$ is the total number of products minus the number of products that are not multiples of $5$:$19900 - 12720 = 7180$.

All possible two-factor products are formed from the numbers $1, 2, 3, 4, \dots, 200$. The number of products out of the total obtained which are multiples of $5$ is:

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